# Is it true that the order of $ab$ is always equal to the order of $ba$?

How do I prove that if $a$, $b$ are elements of group, then $o(ab) = o(ba)$?

For some reason I end up doing the proof for abelian(ness?), i.e., I assume that the order of $ab$ is $2$ and do the steps that lead me to conclude that $ab=ba$, so the orders must be the same. Is that the right way to do it?

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Why on earth are you assuming that the order of $ab$ is $2$? – Chris Eagle Dec 8 '12 at 11:14
This question is related to math.stackexchange.com/questions/225942/… – user26857 Apr 22 '13 at 21:33

Here's an approach that allows you to do some hand-waving and not do any calculations at all. $ab$ and $ba$ are conjugate: indeed, $ba=a^{-1}(ab)a$. It is obvious (and probably already known at this point) that conjugation is an automorphism of the group, and it is obvious that automorphisms preserve orders of elements.

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Calling this «hand-waving» is quite misguided! – Mariano Suárez-Alvarez Apr 22 at 7:33

Hint: Suppose $ab$ has order $n$, and consider $(ba)^{n+1}$.

Another hint is greyed out below (hover over with a mouse to display it):

Notice that $(ba)^{n+1} = b(ab)^na$.

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If $(ab)^n=e$ then $(ab)^na=a$. Since $(ab)^na=a(ba)^n$, $(ba)^n=e$. This proves that the order of $ba$ divides the order of $ab$. By symmetry, the order of $ab$ divides the order of $ab$. Hence the order of $ab$ and the order of $ba$ coincide.

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I think the OP should note that the orders of $a$ and $b$ are both finite. – Babak S. Nov 15 '12 at 20:58
@BabakSorouh Why? The order of ab may be finite while those of a and b are infinite. – Did Nov 15 '12 at 21:11
@Did. I would like to know of a specific example of a group with elements a and b with infinite order but the order of ab is finite. I am not disputing what you are stating. I am a beginning student of algebra and I need " a good stock of examples". I can see that if a and b are inverses of each other then ab=e, but I was hoping for an example where a does not equal b inverse. – Geoffrey Critzer Jun 5 '15 at 10:28
@GeoffreyCritzer Try $b=a^{-1}$. – Did Jun 5 '15 at 10:33
@GeoffreyCritzer In the plane, try $a$ the translation by $(1,0)$ and $ba$ the symmetry $(x,y)\mapsto(y,x)$. – Did Jun 5 '15 at 10:44

By associativity, $(ab)^p=a(ba)^{p-1}b$ for $p\geqslant 1$. If $(ab)^p=e$ then $a(ba)^{p-1}b=e$, so $a(ba)^p=a$ and $(ba)^p=e$. We conclude that for $p\geqslant 1$, $$(ab)^p=e\Leftrightarrow (ba)^p=e.$$

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